Calculating $\int\limits_0^{ + \infty } {{{x\,{\mathop{\ln x\,\rm{d}x}\nolimits} } \over {{{\left( {{x^2} + 1} \right)}^2}}}} $ When i find limit to calculate this, it hard to find. Can anyone finish this ?
$$\int_0^{ + \infty } {{{x\,{\mathop{\ln x\,\mathrm{d}x}\nolimits} } \over {{{\left( {{x^2} + 1} \right)}^2}}}} $$
 A: This is not an answer to the question but a comment on Sami Ben Romdhane's answer, but it is too long for a comment.
Although the integral is $0$ as shown by the hint in Sami's answer, I commented that since the integral from $0$ to $1$ is the negative of the integral from $1$ to $\infty$, it would be interesting to compute the integral from $0$ to $1$. Here is my first attempt which uses
$$
\int_0^1\log(x)x^k\,\mathrm{d}x=-\frac1{(k+1)^2}
$$
to get
$$
\begin{align}
\int_0^1\frac{x\log(x)}{(x^2+1)^2}\,\mathrm{d}x
&=\frac14\int_0^1\frac{\log(x^2)}{(x^2+1)^2}\,\mathrm{d}x^2\\
&=\frac14\int_0^1\frac{\log(x)}{(x+1)^2}\,\mathrm{d}x\\
&=\frac14\int_0^1\log(x)\left(1-2x+3x^2-4x^3+\dots\right)\,\mathrm{d}x\\
&=\frac14\left(-1+\frac12-\frac13+\frac14-\dots\right)\\
&=-\frac{\log(2)}{4}
\end{align}
$$

Interchange of Summation and Integration
It has been suggested that I justify the interchange of summation and integration in the penultimate equation above.
Using the formula
$$
\frac1{(1+x)^2}=\sum_{k=0}^{n-1}(-1)^k(k+1)x^k+(-1)^n\frac{(n+1)x^n+nx^{n+1}}{(1+x)^2}
$$
we see that on $[0,1]$ the difference between $\frac1{(1+x)^2}$ and $\sum\limits_{k=0}^{n-1}(-1)^k(k+1)x^k$ is less than $(2n+1)x^n$. Therefore, the difference between
$$
\int_0^1\frac{\log(x)}{(1+x)^2}\,\mathrm{d}x
$$
and
$$
\int_0^1\log(x)\left(\sum_{k=0}^{n-1}(-1)^k(k+1)x^k\right)\,\mathrm{d}x
$$
is less than
$$
(2n+1)\int_0^1|\log(x)|x^n\,\mathrm{d}x=\frac{2n+1}{(n+1)^2}\to0
$$
A: Hint
By the change variable $t=\frac{1}{x}$ prove
$$\int_0^{ + \infty } {{{x{\mathop{\rm \ln x dx}\nolimits} } \over {{{\left( {{x^2} + 1} \right)}^2}}}} =0$$
A: As the denominator contains $1+x^2,$ let us put $x=\tan\theta $
$$I=\int_0^{\infty } {{{x\ln x\;\mathrm{dx} } \over {{{\left( {{x^2} + 1} \right)}^2}}}}=\frac12\int_0^{\frac\pi2}\sin2\theta \cdot \ln\tan\theta d\theta $$
As $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$$I=\frac12\int_0^{\frac\pi2}\sin2\left(0+\frac\pi2-\theta\right)\cdot \ln\tan\left(0+\frac\pi2-\theta\right) d\theta$$
$$ =-\frac12\int_0^{\frac\pi2}\sin2\theta\cdot\ln\tan\theta d\theta=-I$$
As $\sin2\left(0+\frac\pi2-\theta\right)=\sin(\pi-2\theta)=\sin2\theta$ and $\tan\left(0+\frac\pi2-\theta\right)=\cot\theta=(\tan\theta)^{-1}$
A: You can find the indefinite integral using integration by parts:
$$\begin{align}u&=\ln(x) \\
dv&=\frac{x}{(x^2+1)^2}dx\end{align}$$
$$\begin{align}du&=\space\space\space\frac{1}{x}dx \\
v&=-\frac{1}{2(x^2+1)}\end{align}$$
$$\int {{{x\ln x\;\mathrm{dx} } \over {{{\left( {{x^2} + 1} \right)}^2}}}}=-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int{\frac{1}{(x^2+1)x}}dx$$
The final integral can be solved using partial fraction decomposition:
$$\frac{1}{(x^2+1)x}=-\frac{x}{x^2+1}+\frac{1}{x}$$
A: For completeness here, I should mention that the integral in question is easily amenable to analysis via the residue theorem.  In this case, one would use a keyhole contour about the positive real axis, about which we would consider
$$\oint_C dz \frac{z \, \ln^2{z}}{(1+z^2)^2}$$
The contour integral vanishes on the outer and inner  circular arcs of the keyhole  contour, so we are left with the integrals up the real axis at $\arg{z}=0$, and back along the real axis at $\arg{z}=2 \pi$.  Thus, the contour integral is equal to
$$\int_0^{\infty} dx \frac{x \left [ \ln^2{x}-(\ln{x}+i 2 \pi)^2\right ]}{(1+x^2)^2} = -i 4 \pi \int_0^{\infty} dx \frac{x \, \ln{x}}{(1+x^2)^2} + 4 \pi^2 \int_0^{\infty} dx \frac{x}{(1+x^2)^2}$$
Then again, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the integrand at its poles inside $C$.  The poles re at $z=i-e^{i \pi/2}$ and $z=-i=e^{i 3 \pi/2}$.  For now, I will simply state that the sum of the residues of $z \ln{z}/(1+z^2)^2$ at these poles is $-i \pi$; therefore, the contour integral is $i 2 \pi (-i \pi) = 2 \pi^2$.  Because this is real, the imaginary part of the integral must be zero; therefore the integral
$$\int_0^{\infty} dx \frac{x \, \ln{x}}{(1+x^2)^2} = 0$$
As a bonus, we also see that, equating the real parts of both sides,
$$\int_0^{\infty} dx \frac{x}{(1+x^2)^2} = \frac12$$ 
